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Ex 12.3, 10 - Area of equilateral triangle ABC is 17320.5 cm2 - Area of combination of figures : sector based

  1. Chapter 12 Class 10 Areas related to Circles
  2. Serial order wise
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Ex 12.3, 10 The area of an equilateral triangle ABC is 17320.5 cm2 . With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region. (Use π= 3.14 and √3= 1.73205) Area of shaded region = Area of equilateral triangle – Area of sector ADE – Area of sector BDF – Area of sector CFE It is given that, Δ ABC is an equilateral triangle, So, sides of an triangle = AB = BC = AC Also, Area of an equilateral triangle = 17320.5 cm2 √3/4 (side)2 = 17320.5 (Side)2 = (17320.5 × 4)/√3 (Side)2 = (17320.5 × 4)/1.73205 (Side)2 = (17320.5 × 4× 100000)/173205 (Side)2 = (173205 × 4× 10000)/173205 (Side)2 = 40000 Side = √40000 Side = √(4 ×10000) Side = √((2)^2 ×(100)^2 ) Side = 200 cm Given Radius of each circle = Half length of side of triangle = 200/2 = 100 cm We know that equilateral triangle has all angles 60° ∠ ABC = ∠ BAC = ∠ ACB = 60° So, θ = 60° Area of sector ADE = 𝜃/(360°) 𝜋𝑟2 = (60°)/360 × 3.14×(100)2 = 1/6×3.14×100×100 = 1/3×3.14×50×100 = 15700/3 cm2 Area of sector DBF & CFE have the same value of radius and angle Hence , Area sector ADE = Area sector DBF = Area sector CFE = 15700/3 cm2 Area of shaded region = Area of equilateral triangle – Area of sector ADE – Area of sector BDF – Area of sector CFE = 17320.5 – 15700/3 – 15700/3 – 15700/3 = 17320.5 – 3 × 15700/3 = 17320.5 –15700 = 1620.5 cm2 Hence, area of shaded region = 1620.5 cm2

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